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人気のある微分積分 >

integral from 0 to 1 of sqrt((1+x^2))

解

\int_{0}^{1} (1 + x^{2}) d x

解

\frac{2 + ln ( 1 + 2 )}{2}

+1

十進法表記

1.14779 \dots

解答ステップ

\int_{0}^{1} 1 + x^{2} d x

三角関数による置換を適用する

= \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

積分換算を適用する

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

簡素化

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2) : [\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

限度を計算する:

\frac{1}{2}

= \frac{1}{2} + \frac{1}{2} ln (1 + 2)

簡素化

= \frac{2 + ln ( 1 + 2 )}{2}

グラフ

人気の例

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